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PROBLEM: The figure above shows an accelerometer, a device for measuring the horizontal acceleration of cars and airplanes. A ball is free to roll on a parabolic track described by the equation $y=x^2$, where both x and y are in meters. A scale along the bottom is used to measure the ball’s horizontal position x. Find an expression that allows you to use a measured position x (in m) to compute the acceleration $a_x$ (in $m/s^2$). (For example, $a_x = 3x$ is a possible expression.)

PROFESSOR'S EXPLANATION: By looking at a small portion of the parabola, the balls motion can be approximated to a ramp, with a base of dx, a height of dy, and an angle of $\theta$. The two forces acting on the ball are $F_g$ and the normal force, n. The normal force can be split into a horizontal component parallel to the base of the ramp, and a vertical component parallel to the height of the ramp. The horizontal component equals $n*sin(\theta)$, and the vertical component equals $n*cos(\theta)$.

The next step is what confuses me. He says that $F_g = n*cos(\theta)$. Thus $mg = n*cos(\theta)$ and $n = \frac{mg}{cos(\theta)}$. This implies the sum of the forces vertically is 0, which is confusing because the ball is accelerating down the ramp. This also conflicts with what I've learned before about motion down a ramp, where the normal force is usually $mg * cos(\theta)$

If you continue with his explanation however, everything else makes sense and you reach the correct answer. By substituting what you got for n into the horizontal component of the normal force, you get $F_{x net} = \frac{mgsin(\theta)}{cos(\theta)} = mgtan(\theta)$. This leads to $ma_x = mgtan(\theta)$ which leads to $a_x = gtan(\theta) = g\frac{dy}{dx}$. Looking at the original problem and our theoretical ramp, $\frac{dy}{dx} = 2x$. Thus $a_x = g2x$.

Why is the normal force in this problem $\frac{mg}{cos(\theta)}$ and not $mg * cos(\theta)$ like it normally is? How can you set $F_g$ equal to $n*cos(\theta)$ when the ball is accelerating down the ramp? Any help is appreciated.

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I believe your confusion comes from thinking the ball is accelerating down the ramp. The horizontal acceleration is measured when the ball is in equilibrium with the parabolic block. The ball is actually stationary with respect to the block. Looking at the $m\ a=\text{$\Sigma $F}$ in the y direction:

$m\ a_y=m\ g - n \cos (\theta )$

but in equilibrium $a_y$ is zero and we have your professor's first equation from which we can calculate $n$.

In equilibrium the $x$ acceleration of the ball is the same as the $x$ acceleration of the block and we have

$m\ a_x=n \sin (\theta )$

and substituting for $n$ gives your other equation.

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  • $\begingroup$ When you say block what exactly are you referring to? The only two objects I can get from the problem are the ball and the track, and I don’t see how the ball can be in equilibrium with the track. $\endgroup$
    – ninjaz39
    Oct 19 '20 at 21:27
  • $\begingroup$ Looks like a block with a parabolic track to me. Of course the ball is in equilibrium with the track. With zero acceleration the ball will sit at the bottom of the track. When the block(track) moves with constant acceleration to the right the ball will move up and to the left with respect to the track to an equilibrium position that depends on the acceleration. There will be some wobbling about the equilibrium for a time, but that will damp out with any real material. As long as the acceleration remains constant, the ball will stay in that equilibrium position. $\endgroup$
    – Bill Watts
    Oct 19 '20 at 22:54
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If I understand the question correctly, the figure below may help with some of the geometry. The key idea is that when you have non-zero horizontal acceleration, the ball will move up the curve until it stops at a certain location. Seeing that the ball is in a fixed position relative to the curve/accelerometer and the geometry figure below should allow you to work out the rest.

enter image description here I hope this helps.

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    $\begingroup$ The sum of the horizontal forces is not zero. The ball is accelerating at the same rate as the accelerometer. If the sum of horizontal forces is zero, the ball will be resting at the bottom of the parabola. I don't think working in an accelerating reference frame is useful here. $\endgroup$
    – Mark H
    Oct 19 '20 at 3:33

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